Generalized vectors. Eigenvalues/Eigenvectors.

Let [itex]A \in M_{22} (\mathbb{R})[/itex] with one single eigenvalue λ and one single eigenvector v. We denote w the generalized vector such that [itex](A - λI)w = v[/itex]. Prove that v and w are linearly independent.

2. Relevant equations

I know that if A has only one eigenvalue λ and one eigenvector v that the equation Av = λv is satisfied. That is, (A - λI)v = 0.

3. The attempt at a solution

I thought about this a bit, but I'm having trouble getting this one going. I thought about letting B = (A - λI) so that we get two equations :

Bv = 0 and Bw = v

Then I thought that it could be broken down into two cases, one where λ = 0 and one where λ ≠ 0, but I'm not sure this is the right path to take.

Let [itex]A \in M_{22} (\mathbb{R})[/itex] with one single eigenvalue λ and one single eigenvector v. We denote w the generalized vector such that [itex](A - λI)w = v[/itex]. Prove that v and w are linearly independent.

2. Relevant equations

I know that if A has only one eigenvalue λ and one eigenvector v that the equation Av = λv is satisfied. That is, (A - λI)v = 0.

3. The attempt at a solution

I thought about this a bit, but I'm having trouble getting this one going. I thought about letting B = (A - λI) so that we get two equations :

Bv = 0 and Bw = v

Then I thought that it could be broken down into two cases, one where λ = 0 and one where λ ≠ 0, but I'm not sure this is the right path to take.

Let [itex]A \in M_{22} (\mathbb{R})[/itex] with one single eigenvalue λ and one single eigenvector v. We denote w the generalized vector such that [itex](A - λI)w = v[/itex]. Prove that v and w are linearly independent.

2. Relevant equations

I know that if A has only one eigenvalue λ and one eigenvector v that the equation Av = λv is satisfied. That is, (A - λI)v = 0.

That's true for any A such that [itex]\lambda[/itex] is an eigenvalue with eigenvector v. The fact that A has only that eigenvalue and one independent eigenvector, tells you that there exist a vector u such that [itex]A- \lambda I)u\ne 0[/itex] but that [itex](A- \lambda I)^2u= 0[/itex].

(It is never true that a linear operator has "one single eigenvector" because any multiple of an eigenvector is an eigenvector.)

3. The attempt at a solution

I thought about this a bit, but I'm having trouble getting this one going. I thought about letting B = (A - λI) so that we get two equations :

Bv = 0 and Bw = v

Then I thought that it could be broken down into two cases, one where λ = 0 and one where λ ≠ 0, but I'm not sure this is the right path to take.